Question 4 of 12

Question 4 (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. (a) $f(n)=\sqrt{n^3+1}$, proposed domain $\mathbb{Z}$, codomain $\mathbb{R}$. (b) $f(x)=\sqrt{x^4+1}$, domain $\mathbb{R}$. (c) $f(x)=2x^4+1$, $\mathbb{R}\to\mathbb{R}$. (d) $f(x)=x^3+2$ and $(f+g)(x)=x^3+x+2$, both $\mathbb{R}\to\mathbb{R}$.

Find. (a) Whether $f$ is a valid function $\mathbb{Z}\to\mathbb{R}$. (b) The range of $f$. (c) Whether $f$ is one-to-one. (d) $g(x)$, then $(g\circ f^{-1})^{-1}(x)$.

Approach. (a) check that $n^3+1\ge 0$ for every integer $n$ (needed for the square root to be real); (b) minimize $x^4+1$ then take square roots; (c) look for two distinct inputs with equal outputs; (d) subtract $f$ from $f+g$ to isolate $g$, then use $g=\text{id}$ to simplify the composite inverse.

1. (a) Is $f(n)=\sqrt{n^3+1}$ a function $\mathbb{Z}\to\mathbb{R}$? A function must assign a well-defined real value to every element of its stated domain. Test $n=-2\in\mathbb{Z}$: $n^3+1=(-2)^3+1=-7\lt 0$, so $\sqrt{-7}$ is not a real number — $f(-2)$ is undefined. Since the rule fails to produce an output for at least one integer, it is not total on $\mathbb{Z}$: $$\boxed{f(n)=\sqrt{n^3+1}\text{ is NOT a function from }\mathbb{Z}\text{ to }\mathbb{R}\text{ (fails at }n\le -2\text{, e.g. }n=-2\text{)}}$$
2. (b) Range of $f(x)=\sqrt{x^4+1}$, $x\in\mathbb{R}$. Since $x^4\ge 0$ for every real $x$, the expression under the root satisfies $x^4+1\ge 1$, with the minimum $x^4+1=1$ attained exactly at $x=0$. As $|x|\to\infty$, $x^4+1\to\infty$. The square root is continuous and increasing on $[1,\infty)$, so it maps this interval onto $[1,\infty)$ as well: $$\min f = \sqrt{0+1}=1\ (\text{at }x=0),\qquad f(x)\to\infty\text{ as }|x|\to\infty$$ $\boxed{\text{Range}=[1,\infty)}$
3. (c) Is $f(x)=2x^4+1$ one-to-one on $\mathbb{R}$? $f$ is an even function ($f(-x)=2(-x)^4+1=2x^4+1=f(x)$), so any nonzero input and its negative give the same output. Concretely, $f(1)=2(1)^4+1=3$ and $f(-1)=2(-1)^4+1=3$, so $f(1)=f(-1)$ with $1\ne -1$: $$\boxed{f\text{ is NOT one-to-one (counterexample: }f(1)=f(-1)=3\text{)}}$$
4. (d) Find $g(x)$, then $(g\circ f^{-1})^{-1}(x)$. Since $(f+g)(x)=f(x)+g(x)$, subtract: $g(x)=(f+g)(x)-f(x)=(x^3+x+2)-(x^3+2)=x$. So $g$ is the identity function, $g(x)=x$. For the identity function, $g\circ h=h$ for any $h$ (composing with identity changes nothing), so: $$(g\circ f^{-1})(x) = g(f^{-1}(x)) = f^{-1}(x)$$ Taking the inverse of both sides, $\big(g\circ f^{-1}\big)^{-1}=\big(f^{-1}\big)^{-1}=f$: $$\boxed{(g\circ f^{-1})^{-1}(x) = f(x) = x^3+2}$$